If `(1+x+x^(2))^(20) = a_(0) + a_(1)x^(2) "……" + a_(40)x^(40)`, then following questions.
The value of `a_(0) + a_(1) + a_(2) + "……" + a_(19)` is

Answer

B

Explanation

Let, `(1+x+x^(2))^(20) = underset(r=0)overset(40)suma_(r)x^(r ) " "(1)`
Replacing x by `1//x`, we get
`(1+x1/x+1/(x^(2)))^(20) = underset(r=0)overset(40)suma_(r)(1/x)^(r )`
or `(1+x+x^(2))^(20) = underset(r=0)overset(40)suma_(r)x^(40-r) " "(2)`
Since (1) and (2) are same series, coefficient of `x^(r )` is (1) `=` coefficient of `x^(r)` in (2).
`rArr a_(r) = a_(40-r)`
In (1) Putting `x = 1`, we get
`3^(20) = a_(0)+a_(1)+a_(2)+"...."+a_(40)`
`= (a_(0)+a_(1)+a_(2)+"...."+a_(19))+a_(20)+(a_(21)+a_(n+2)+"..."+a_(40))`
`= 2(a_(0)+a_(1)+a_(2)+"...."+a_(19))+a_(20)" "( :' a_(r) = a_(40-r))`
or `a_(0) + a_(1) + a_(2) + "......."+ a_(19) = 1/2 (3^(20)-a_(20)) = 1/2(9^(10) - a_(20))`
Also,
`a_(0)+3a_(1)+5a_(2)+81a_(40)`
`= (a_(0)+81a_(40))+(3a_(1)+79a_(39))+"...."+(39a_(19)+43a_(21))+41a_(20)`
`= 82(a_(0) + a_(1) + a_(2) + "......" + a_(19)) + 41a_(20)`
`= 41 xx 3^(20)`
`a_(0)^(2) - a_(1)^(2) + a_(2)^(2) - a_(3)^(2) + "....."` suggests that we have to multiply the two expansions.
Replacing x by `-1//x` in (1), we get
`(1-1/x+1/(x^(2)))^(20) = a_(0) - (a_(1))/(x)+(a_(2))/(x^(2))-"...."+(a_(40))/(a_(40))`
`rArr (1-x+x^(2))^(20) = a_(0)x^(40) - a_(1)x^(39) + a_(2)x^(38) - "....."a_(40)" "(3)`
Clearly,
`a_(0)^(3) - a_(1)^(2) + a_(2)^(2) + "....."+ a_(0)^(2)` is the coefficeint of `x^(40)` in `(1+x+x^(2)) (1-x+x^(2))^(20)`
= Coefficient of `x^(40)` in `(1+x^(2)+x^(4))^(20)`
In `(1+x^(2)+x^(4))^(20)` replace `x^(2)`, by y, then the coefficientof `y^(20)` in `(1+y+y^(2))^(20)` is `a_(20)`.
Hence `a_(0)^(2) - a_(1)^(2) -"......"+a_(40)^(2) = a_(20)`
or `(a_(0)^(2) - a_(1)^(2) + a_(2)^(2) - "....." - a_(19)^(2)) + a_(20)^(2) + (-a_(21)^(2) + "....." + a_(40)^(2)) = a_(20)`
or `2(a_(0)^(2) - a_(1)^(2) + a_(2)^(2) - "....." - a_(19)^(2)) + a_(20)^(2) = a_(20)`
or `a_(0)^(2) - a_(1)^(2) -"......" - a_(19)^(2) = (a_(20))/(2)[1-a_(20)]`